MATH2321, Calculus III for Science and Engineering, Fall Name (Printed) Print your name, the date, and then sign the exam on the line

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1 MATH2321, Calculus III for Science and Engineering, Fall Exam 2 Name (Printed) Date Signature Instructions STOP. above. Print your name, the date, and then sign the exam on the line This exam consists of 5 problems, each worth 2 points apiece, for a total of 1 points. Work as many problems as you can within 65 minutes. You may work the problems in any order, so use your time wisely. If you finish early, I encourage you to check your work before you hand in the exam. Work each problem in the space provided, or on the back of the preceding page. Be sure to indicate if your work runs to the back of a page. Circle or box your final answer to each problem, and show all work. I will assign partial credit for incorrect answers based upon the work that you submit. You may use your calculator on this exam. However, answers from your calculator without supporting work are worth zero points. The use of any other electronic device is NOT permitted. This exam is closed book. If you brought your textbook or any notes with you today, keep them out of sight for the duration of the exam. If the wording of any problem is unclear, raise your hand. I will come to your desk and attempt to clarify. Do not turn the page until I give the signal to begin. Good luck!

2 MATH2321, Calculus III for Science and Engineering, Fall Problem 1. [2 points] You must build a rectangular shipping crate with volume 2 ft 3. The sides cost $1/ft 2, the top costs $2/ft 2, and the bottom costs $3/ft 2. What dimensions minimize the cost of the crate? Answer: We first sketch the crate in Figure 1, with dimensions The volume V (a, b, c) of the crate in ft 3 then satisfies b ft c ft a ft. (1) V (a, b, c) abc 2, (2) so V (a, b, c) is the constraint function in this problem. Similarly, the cost C(a, b, c) of the crate in dollars is determined in terms of the respective areas A top, A bottom, and A sides of the top, bottom, and four side panels as C(a, b, c) 2 A top + 3 A bottom + A sides, 2 (bc) + 3 (bc) + (ab + ac + ab + ac), 5 bc + 2 ab + 2 ac. (3) Since we seek to minimize the cost, C(a, b, c) is the objective function in this problem. a c b Figure 1: Rectangular shipping crate. To minimize C(a, b, c) subject to the constraint V (a, b, c) 2, we apply the method of Lagrange multipliers and solve the vector equation C(a, b, c) λ V (a, b, c), (4)

3 MATH2321, Calculus III for Science and Engineering, Fall or in components, C a λ V a, C b λ V b, (5) C c λ V c, where the Lagrange multiplier λ is an undetermined constant. Evaluating the partial derivatives with respect to (a, b, c) in (5), we find the equations 2 b + 2 c λ b c, 5 c + 2 a λ a c, 5 b + 2 a λ a b. To solve the equations in (6), we follow our nose and simply try to solve each equation in turn. Thus, the first equation in (6) implies 2 b + 2 c λ b c, (6) λ 1 (2 b + 2 c), bc (7) λ 2 c + 2 b. Substituting for λ in the second equation of (6), we find ( 2 5 c + 2 a c + 2 ) a c, b 5 c + 2 a 2a + 2 ac, b 5 c 2 ac b. Dividing the latter equation on both sides by c, we obtain (8) 5 2 a b a 5 2 b. (9) We now use the results in (7) and (9) to substitute for both λ and a in the third equation of (6), ( ) ( b b 2 c + 1 ) ( ) 5 b 2 b b, 1 b 5 b2 c + 5 b, 5 b 5 b2 c, (1) 1 b c,

4 MATH2321, Calculus III for Science and Engineering, Fall so c b. (11) Note that the relation in (11) says that the top and the bottom of the crate must be square. A square is the special rectangle which has the minimum perimeter for a given area. To determine b, we finally substitute a 5/2 b and c b into the constraint equation V (a, b, c) abc 2 to obtain Thus and the crate has dimensions Alternate Solution: 5 2 b3 2, b 3 8, b 2. (12) b 2, c 2, a 5, (13) 2 ft 2 ft 5 ft. (14) Eliminate the variable a by solving the constraint a 2 bc. (15) The cost C is then given as a function of b and c by f(b, c) C(a, b, c) 5bc + 4 a2/bc c + 4 b. (16) Because b and c are constrained only by the bounds b, c, we minimize f by solving for its critical points, where f b f c Substituting for c in the second equation, 5c 4 b 2 c 8 b 2, 5b 4 c 2 b 8 c 2. (17) b 8 (8/b 2 ) 2 b4 8 b ( b 3 8 ). (18) The non-zero solution is b 2, from which c 8/b 2 2 also, and a is determined by the constraint to be a 5, exactly as before.

5 MATH2321, Calculus III for Science and Engineering, Fall Problem 2. [2 points] Evaluate the iterated integral I y 2 sin ( x 2 + y 2) dx dy. y/ 3 Answer: The inner integral of sin(x 2 + y 2 ) with respect to x cannot be evaluated analytically. Instead, to evaluate the iterated integral, we rewrite I as an integral over a region R in the xy-plane, I sin(x 2 + y 2 ) da. (19) Here the region R is bounded by the curves with equation R x min y 3, x max 16 y 2. (2) The equation for x min describes a line through the origin with slope m 3, and the equation for x max describes the semi-circle with radius r 4 in the positive-x half-plane. See Figure 2. The line and the semi-circle intersect where x 2 + y 2 x 2 + 3x 2 4x 2 16 x 2, y 2 3. (21) The integration limits y min and y max 2 3 then imply that R is the wedge-shaped region in Figure 2. Because R is a wedge, we naturally try to evaluate I as an iterated integral in polar coordinates. In these coordinates, I θmax 4 sin(r 2 ) r dr dθ, (22) where we remember to use the polar area element da rdrdθ. By trigonometry, the angle θ max is related to the slope m 3 of the line by tan θ max 3 θ max π 3. (23) The polar integral can be evaluated with the substitution u r 2, du 2rdr, after which π/ I 2 sin u du dθ π [ ] 16 cos u 6. (24) Hence I π (1 cos(16)) (25) 6

6 MATH2321, Calculus III for Science and Engineering, Fall sqrt(3) yx sqrt(3) R θ max r4 2 Figure 2: Region R in the xy-plane. Problem 3. [2 points] Consider the solid region S in space which is bounded by the surfaces y x 2, x y 2, z, and z x + y, with all coordinates measured in meters. Suppose that S is composed of a material with density Compute the mass of S. µ(x, y, z) xy kg/m 3. Answer: The key to this problem is to understand the geometry of S, through the geometry of the bounding surfaces y x 2, x y 2, z, z x + y. (26) Of course, z describes the xy-plane. Let us next think about the surfaces described by the equations y x 2 and x y 2. Since neither equation involves z, the surfaces in space will by cylinders, aligned with the z-axis, over the corresponding parabolas in the xy-plane. These parabolas are graphed in Figure 3.

7 MATH2321, Calculus III for Science and Engineering, Fall y y x 2 x y 2 1 R 1 x Figure 3: Intersecting parabolas. The region in space between the surfaces y x 2 and x y 2 is a vertical cylinder over the region R shaded in green in Figure 6. To describe the solid S itself, we slice the cylinder at z to form the lower boundary for S. The upper boundary for S must then be given by the plane z x + y. Note that the the plane z x + y passes through the origin and indeed slants upwards over the positive quadrant in the xy-plane, since z increases with x and y. With this description for S, the integral which computes the mass M of S is straightforward to write, M µ(x, y, z) dv, S [ zmaxx+y ] xy dz da, (27) R z min xmax ymax x+y x min y min xy dz dy dx,

8 MATH2321, Calculus III for Science and Engineering, Fall for some y min, y max, x min, and x max appropriate to describe the integral over the region R in Figure 6. To determine y min and y max, we just consider the equations for the respective lower and upper boundaries of R. The lower boundary of R is given by y x 2, so we set y min x 2. Similarly, the upper boundary of R is described by the equation x y 2, so solving for y, we set y max x. As for the limits x min and x max, we see that the first vertical segment enters R at x min, and the last vertical segment leaves R at the point where the parabolas y x 2 and x y 2 intersect. At this point, x x 2, so x max 1. Altogether, the iterated integral which computes the mass M of S becomes M We now integrate in stages to find M 1 x x+y 1 x x 2 1 x 1 x 1 1 So finally we compute the mass to be x 2 [ ] x+y xyz dy dx, x 2 x y (x + y) dy dx, ( x 2 y + x y 2) dy dx, x 2 [ 1 2 x2 y x x y3] dx, x ( x x5/2 1 2 x6 1 3 x7 [ 1 8 x x7/ x x8 M , xy dz dy dx. (28) ) dx, ] 1. (29) kg. (3) Problem 4. [2 points] Consider a large spherical tank, of radius 5 m. The tank is filled with water to a height of 2 m, as shown in Figure 4. What is the volume of water in the tank?

9 MATH2321, Calculus III for Science and Engineering, Fall φ min 5 m 3 m ψ water 4 m 2 m Figure 4: Spherical tank containing water. Answer: The volume of water in the tank can be determined by integrating in either spherical or cylindrical coordinates. Let us start with spherical coordinates, in which case the volume V of water is given by an integral of the form V θmax φmax ρmax θ min φ min ρ min ρ 2 sin φ dρ dφ dθ. (31) Since the region of integration is symmetric under rotations around the z-axis, we see directly that θ min, θ max 2π. (32) Otherwise, we integrate first over ρ, in radial rays extending outwards from the center of the sphere to the boundary where ρ max 5. (33) Otherwise, ρ min is determined by the equation for the surface of the water in the tank. Assuming that the origin of our coordinate system coincides with the center of the tank, we see that the surface of the water occupies the horizontal plane z 3 ρ cos φ 3. (34) Hence the equation for the surface of the water in spherical coordinates is ρ min 3 cos φ. (35)

10 MATH2321, Calculus III for Science and Engineering, Fall Finally we must determine the azimuthal angles φ min and φ max. Clearly φ max π. (36) Otherwise, the supplementary angle ψ in the figure is determined by trigonometry to satisfy cos ψ 3 5. (37) Hence φ min satisfies cos φ min 3 5. (38) As will be seen, when we integrate, all we really need to know is the cosine of φ min, not φ min itself. We are left to compute the volume integral V We first integrate over ρ to obtain 2π π 5 φ min 3/ cos φ ρ 2 sin φ dρ dφ dθ. (39) V 1 3 2π π φ min 2π π [ 1 φ min 3 ρ3 ] 5 3/ cos φ sin φ dφ dθ, ( ) cos 3 sin φ dφ dθ. φ The integral over φ is also elementary. Note that the integral of the term proportional to sin φ/cos 3 φ can be performed after a substitution u cos φ, since du sin φ dφ. Thus V 2π [ 125 cos φ + 27 ] 1 π 3 2 cos 2, φ φ min 2π [( ) ( cos φ min 27 )] (41) cos 2. φ min (4) Since the expression for V involves only cos φ min 3/5, we substitute to obtain V 52π m 3. (42) Alternate Solution: As a check, let us also compute the volume of water using cylindrical coordinates. In this case, the volume integral takes the form V 2π rmax zmax r min z min r dz dr dθ. (43)

11 MATH2321, Calculus III for Science and Engineering, Fall The limits of integration for z are again determined by the equations for the curved boundary of the sphere and the flat surface of the water. The equation for a sphere of radius 5 which is centered at the origin is x 2 + y 2 + z 2 25, or in cylindrical coordinates, r 2 + z 2 25 z min 25 r 2. (44) I take the negative square-root because z min is clearly negative in the figure. On the other hand, the flat surface of the water lies in the horizontal plane z max 3. (45) As for the radial limits, clearly r min. To determine r max, we can either use trigonometry or algebra. Via algebra, we solve the condition for the semicircle z 25 r 2 to intersect the line z 3, ie r 2 r max 4. (46) The volume integral in cylindrical coordinates becomes V 2π r 2 r dz dr dθ. (47) We now evaluate exactly as before. V 2π 2π 4 52π 3, r [ 3 + ] 25 r 2 dr, [ 3 2 r2 1 3 ( 25 r 2 ) 4 3/2], (48) Problem 5. [2 points] Let M be the surface in R 3 which is parametrized by r(u, v) ( (1 + u) cos v, (1 + u) sin v, v ), for parameter ranges u 1 and v 6π. Compute the surface area element ds in terms of u and v, and write an iterated integral which gives the total surface area of M. You do NOT have to calculate the value of the integral.

12 MATH2321, Calculus III for Science and Engineering, Fall Answer: To compute the surface area element ds we use the general formula ds r u r v du dv. (49) The partial derivatives of r are given by r u (cos v, sin v, ), r v ( (1 + u) sin v, (1 + u) cos v, 1 ). (5) So the cross-product is i j k r u r v cos v sin v (1 + u) sin v (1 + u) cos v 1, i sin v j cos v + k [ cos 2 v (1 + u) + sin 2 v (1 + u) ], i sin v j cos v + k (1 + u). (51) The magnitude of the cross-product is finally r u r v sin 2 v + cos 2 v + (1 + u) (1 + u) 2, (52) and ds 1 + (1 + u) 2 du dv. (53) The total surface area of M is then given by the integral of ds over the given ranges for the parameters u and v, 6π 1 1 SA 1 + (1 + u) 2 du dv 6π 1 + (1 + u) 2 du. (54) Though not required, the surface area can be evaluated analytically with a tricky integration formula, 1 + (1 + u) 2 du 12 1 (1 + u) + (1 + u) [ ] ln 1 + u (1 + u) 2. As a result, SA 6π (55) [ 5 1 ( + 2 ln 2 + ) (1 2 2 ln + 2) ] (56) More interesting is the picture of the surface M itself. A computer graph of M appears in Figure 5 on the next page. Before turning the page, can you guess what M looks like?

13 MATH2321, Calculus III for Science and Engineering, Fall Figure 5: Graph of the parametrized surface M in space. M is a helical ramp, as one might see in a parking garage.